Deriving Sines of 30, 45, 60 and 90 Degrees

Date: 06/02/99 at 09:41:14
From: Michael
Subject: Where does the value of sin(pi/3) come from?
To Dr. Math,
Where do the values for sin 30, sin 60, etc. come from? We're told to
memorize the values and take them for granted, but what is the actual
derivation for getting the values?
In other words, is there any way to prove that sin 30 = 1/2 or that
sin 60 = sqrt(3)/2? Of course, I have the same question about cos,
tan, etc, but if you explain sine I can figure those out.
Thanks,
Mike

Date: 06/02/99 at 10:29:23
From: Doctor Rick
Subject: Re: Where does the value of sin pi/3 come from?
Hi, Mike, thanks for the question!
Surely you weren't just told to take these facts for granted - they
aren't hard at all to derive. Perhaps you were absent the day you were
shown the reason for them.
You didn't mention 45 degrees (sin 45 = pi/4). Do you understand why
sin(pi/4) = sqrt(2)? I'll go over it, because this will lead into the
angles you asked about.
Consider a 45-45-90 triangle. Because 2 angles are equal, it is
isosceles: both legs have the same length. If you know one leg (a),
you know both legs and you can use the Pythagorean Theorem to find the
hypotenuse c.
/|
/ |
c / |
/ | a
/ |
/ |
/____________|
a
c = sqrt(a^2 + a^2)
= sqrt(2*a^2)
= sqrt(2)*a
So these are the proportions of the sides of a 45-45-90 triangle:
B
/|
/ |
sqrt(2) / |
/ | 1
/ |
/ |
/____________|
A 1 C
In other words, the ratio of the hypotenuse to either leg is
sqrt(2):1. In terms of trigonometry,
sin(45) = sin(A)
= BC / AB
= 1/sqrt(2)
= sqrt(2)/2
Now for the 30-60-90 triangle. It isn't isosceles like the 45-45-90
triangle, but it is half of an equilateral triangle:
/|\
/ | \
c / | \ c
/ | \
/ a| \
/ | \
/______|______\
b b
The angles of the equilateral triangle are 60 degrees. The top angle
is bisected, giving the 30 degree angle. And since the sides of the
equilateral triangle are equal, 2b = c. That's the key - the side
opposite the 30 degree angle (b) is half the hypotenuse (c).
If we know b, then we know c = 2b, and we can fill in the Pythagorean
Theorem:
(2b)^2 = a^2 + b^2
a = sqrt((2b)^2 - b^2)
= sqrt(3b^2)
= sqrt(3)*b
Now we have the proportions of the sides of a 30-60-90 triangle:
B
/|
/ |
/ |
2 / |
/ | sqrt(3)
/ |
/______|
A 1 C
As before, we can evaluate the sine of angle A, which is 60 degrees:
sin(60) = BC / AB
= sqrt(3)/2
Likewise, for angle B:
sin(30) = AC / AB
= 1/2
It's good to memorize those two proportion figures (or, if you're like
me, remember how to derive them using Pythagoras). Each figure is a
family of facts that will be useful for trigonometry and geometry. I
think this is much better than just memorizing the trig facts in
isolation; the figures tie them all together, and show how to derive
them at the same time.
- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/